@888: "I came across this Underground Sign problem a friend showed me.
<url>
Assume the following,
Concentric circle radii are a and b for inner and outer circles respectively.
The height of the dark blue rectangle is t and length l. The centre of area of 
the rectangle shares the origin of the circles.
Find an expression of the total area of interception of the red and blue areas.
I'll post my 'answer' when i find one. "
    @888: ""Lynn Kurtz" <email_address> wrote in message 
news:<email_address>
| On Sun, 5 Dec 2004 00:34:50 -0000, "mechy engineery" <email_address>
|
| >I came across this Underground Sign problem a friend showed me.
| >
| ><url>
| >
| >Assume the following,
| >
| >Concentric circle radii are a and b for inner and outer circles respectively.
| >
| >The height of the dark blue rectangle is t and length l. The centre of area 
of
| >the rectangle shares the origin of the circles.
| >
| >Find an expression of the total area of interception of the red and blue 
areas.
| >
| >I'll post my 'answer' when i find one.
|
| Assuming you mean the intersection of the rectangle with the annulus,
| it is easy to set up as an integral on y. Note that the length of the
| rectangle is irrelevant, at least as long as it covers the circles.
|
| Area = 4*int(sqrt(b^2 - y^2) - sqrt(a^2 - y^2), y=0..t/2)
|
| This can be done by a trig substitution and the answer is:
|
|  (t/2) (4 b^2  - t^2 )^(1/2)  + 2b^2 arctan( t / sqrt(4b^2 - t^2) )
|
|
| - (t/2) (4 a^2  - t^2 )^(1/2)  - 2a^2 arctan( t / sqrt(4a^2 - t^2) )
|
Hi!
Yes I know it's an integral and having problem setting it up.
I know it's basically finding half the area of one of the intersections then 
multiplying it by 4.
I only got up to this far: "
    @888: ""mechy engineery" <email_address> wrote in message 
news:<email_address>
|
| news:<email_address>
|| On Sun, 5 Dec 2004 00:34:50 -0000, "mechy engineery" <email_address>
||
|| >I came across this Underground Sign problem a friend showed me.
|| >
|| ><url>
|| >
|| >Assume the following,
|| >
|| >Concentric circle radii are a and b for inner and outer circles 
respectively.
|| >
|| >The height of the dark blue rectangle is t and length l. The centre of area
| of
|| >the rectangle shares the origin of the circles.
|| >
|| >Find an expression of the total area of interception of the red and blue
| areas.
|| >
|| >I'll post my 'answer' when i find one.
||
|| Assuming you mean the intersection of the rectangle with the annulus,
|| it is easy to set up as an integral on y. Note that the length of the
|| rectangle is irrelevant, at least as long as it covers the circles.
||
|| Area = 4*int(sqrt(b^2 - y^2) - sqrt(a^2 - y^2), y=0..t/2)
||
|| This can be done by a trig substitution and the answer is:
||
||  (t/2) (4 b^2  - t^2 )^(1/2)  + 2b^2 arctan( t / sqrt(4b^2 - t^2) )
||
||
|| - (t/2) (4 a^2  - t^2 )^(1/2)  - 2a^2 arctan( t / sqrt(4a^2 - t^2) )
||
|
| Hi!
|
| Yes I know it's an integral and having problem setting it up.
|
| I know it's basically finding half the area of one of the intersections then
| multiplying it by 4.
|
| I only got up to this far:
[1] r*sin(theta) = t
[2] int(r*theta, r = a..b)
But theta is arcsin(t/r), therefore,
int(r*arcsin(t/r), r=a..b)
Is this correct?
I'm very interested in your method.
Is y the vertical of the right angle 'triangle'?
If so, how did you derive this "sqrt(b^2 - y^2) - sqrt(a^2 - y^2)""
    @888: ""mechy engineery" <email_address> wrote in message 
news:<email_address>
|
| news:<email_address>
||
|| news:<email_address>
||| On Sun, 5 Dec 2004 00:34:50 -0000, "mechy engineery" <email_address>
|||
||| >I came across this Underground Sign problem a friend showed me.
||| >
||| ><url>
||| >
||| >Assume the following,
||| >
||| >Concentric circle radii are a and b for inner and outer circles
| respectively.
||| >
||| >The height of the dark blue rectangle is t and length l. The centre of area
|| of
||| >the rectangle shares the origin of the circles.
||| >
||| >Find an expression of the total area of interception of the red and blue
|| areas.
||| >
||| >I'll post my 'answer' when i find one.
|||
||| Assuming you mean the intersection of the rectangle with the annulus,
||| it is easy to set up as an integral on y. Note that the length of the
||| rectangle is irrelevant, at least as long as it covers the circles.
|||
||| Area = 4*int(sqrt(b^2 - y^2) - sqrt(a^2 - y^2), y=0..t/2)
|||
||| This can be done by a trig substitution and the answer is:
|||
|||  (t/2) (4 b^2  - t^2 )^(1/2)  + 2b^2 arctan( t / sqrt(4b^2 - t^2) )
|||
|||
||| - (t/2) (4 a^2  - t^2 )^(1/2)  - 2a^2 arctan( t / sqrt(4a^2 - t^2) )
|||
||
|| Hi!
||
|| Yes I know it's an integral and having problem setting it up.
||
|| I know it's basically finding half the area of one of the intersections then
|| multiplying it by 4.
||
|| I only got up to this far:
|
|
| [1] r*sin(theta) = t
|
| [2] int(r*theta, r = a..b)
|
| But theta is arcsin(t/r), therefore,
|
| int(r*arcsin(t/r), r=a..b)
|
| Is this correct?
|
| I'm very interested in your method.
|
| Is y the vertical of the right angle 'triangle'?
|
| If so, how did you derive this "sqrt(b^2 - y^2) - sqrt(a^2 - y^2)"
|
|
I assume "t" is t/2  opss. "
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